If a point $\,(a,b)\,$ is on the graph of a one-to-one function $\,f\,$,
then the point $\,(b,a)\,$ is on the graph of its inverse $\,f^{-1}\,$.
This is because the input/output roles for a function and its inverse are switchedan input to one is an output from the other.

As discussed below, the point $\,(b,a)\,$ is found by reflecting the point $\,(a,b)\,$
about the line $\,y = x\,$,
when graphed in a coordinate system where
‘$\,1\,$’ on the $\,x\,$-axis is the same as
‘$\,1\,$’ on the $\,y\,$-axis.

From this observation, we will see below
that the graph of an inverse function is very easy to obtain if we have the
graph of the original functionjust reflect the original graph about the line
$\,y = x\,$.

Finding a Point $\,(b,a)\,$ from a Given Point $\,(a,b)\,$

Start with a coordinate system where ‘$\,1\,$’ on the $\,x\,$-axis is the same as
‘$\,1\,$’ on the $\,y\,$-axis.
This forces distances measured on either axis to have exactly the same length.
Also, the line $\,y = x\,$ (which has slope $\,1\,$) will be at a $\,45^{\circ}\,$ angle in such a coordinate system.

Plot a point $\,(a,b)\,$.
This is the big green point in the diagram at right.

Locate the number $\,b\,$ on the $\,x\,$-axis, as follows (use the blue arrows):

go from the green point, horizontally, to the line (horizontal blue arrow)

points on the line $\,y = x\,$ have the same $\,x\,$ and $\,y\,$ values

since we know this point has $\,y\,$-value equal to $\,b\,$, its $\,x\,$-value must also be $\,b\,$

mark $\,b\,$ on the $\,x\,$-axis (vertical blue arrow)

Locate the number $\,a\,$ on the $\,y\,$-axis, as follows (use the purple arrows):

go from the green point,vertically, to the line (vertical purple arrow)

points on the line $\,y = x\,$ have the same $\,x\,$ and $\,y\,$ values

since we know this point has $\,x\,$-value equal to $\,a\,$, its $\,y\,$-value must also be $\,a\,$

mark $\,a\,$ on the $\,y\,$-axis (horizontal purple arrow)

Observe that the green shaded region is a square: all sides have length $\,a - b\,$, and all angles are $\,90^{\circ}\,$.
The desired point $\,(b,a)\,$ (large black point) has been located as the corner of the square
opposite the original green point.

When you fold a square along a diagonal, the opposite corners meet.
Thus, the green and black dots will coincide after folding.

Based on these results, simplify the process:
to get from the green point to the black point, just fold along the line $\,y = x\,$.
That is, to get from a point $\,(a,b)\,$ to a point $\,(b,a)\,$, just fold along the line $\,y = x\,$.

Notice that the line $\,y = x\,$ is the perpendicular bisector
of the segment
connecting the original point $\,(a,b)\,$ and the reflected point $\,(b,a)\,$.

The Graph of an Inverse Function

By definition, the graph of a function is the set of all its input/output pairs,
as the inputs vary over the domain.

Thus,
$$\text{the graph of } f^{-1} = \{(x,f^{-1}(x))\ |\ x\in\text{dom}(f^{-1})\}\,$$
Since $\,x\,$ is a dummy variable, this set can also be written using the dummy variable $\,y\,$:
$$\text{the graph of } f^{-1} = \{(y,f^{-1}(y))\ |\ y\in\text{dom}(f^{-1})\}\,$$
Now, we have:

the graph of $\,f^{-1}\,$

$= \{(y,f^{-1}(y))\ |\ y\in\text{dom}(f^{-1})\}$

definition of the graph of a function

$= \{(y,f^{-1}(y))\ |\ y\in\text{ran}(f)\}$

since $\text{ran}(f) = \text{dom}(f^{-1})$

$= \{(f(x),x)\ |\ x\in\text{dom}(f)\}$

$y\in\text{ran}(f)\,$ if and only if there exists $\,x\in\text{dom}(f)\,$ such that $\,f(x) = y\,$;
and, $\,f(x) = y\,$ if and only if $x = f^{-1}(y)$

Thus, the graph of $\,f^{-1}\,$ is precisely the set of points that make up the graph of $\,f\,$, but with
the coordinates switched!
We've already seen that ‘switching coordinates’ is accomplished by
reflecting about the line $\,y = x\,$
in a coordinate system where the scales on the horizontal and vertical axes are
identical.
Thus, we have:

GRAPHING AN INVERSE FUNCTION

Let $\,f\,$ be a one-to-one function, so that it has an inverse $\,f^{-1}\,$.

The graph of $\,f^{-1}\,$ is found as follows:

create a coordinate system where ‘$1$’ on the $x$-axis is the same as
‘$1$’ on the $y$-axis

graph $\,f\,$ in this coordinate system

reflect (‘mirror’) the graph of $\,f\,$ about the line $\,y = x\,$

the reflected graph is the graph of $\,f^{-1}\,$

Several different one-to-one functions and their reflections in the line
$\,y = x\,$ are shown below.

Master the ideas from this section
by practicing the exercise at the bottom of this page.